Largest and smallest number of internal nodes in red-black tree?

Question

The smallest number of internal nodes in a red-black tree with black height of k is 2^k-1 which is one in the following image:

The largest number of internal nodes with black height of k is 2^2k-1 which, if the black height is 2, should be 2⁴ - 1 = 15. However, consider this image:

The number of internal nodes is 7. What am I doing wrong?

Answer 1

(I've completely rewritten this answer because, as the commenters noted, it was initially incorrect.)

I think it might help to think about this problem by using the isometry between red-black trees and 2-3-4 trees. Specifically, a red-black tree with black height h corresponds to a 2-3-4 tree with height h, where each red node corresponds to a key in a multi-key node.

This connection makes it easier for us to make a few neat observations. First, any 2-3-4 tree node in the bottom layer corresponds to a black node with either no red children, one red child, or two red children. These are the only nodes that can be leaf nodes in the red-black tree. If we wanted to maximize the number of total nodes in the tree, we'd want to make the 2-3-4 tree have nothing but 4-nodes, which (under the isometry) maps to a red/black tree where every black node has two red children. An interesting effect of this is that it makes the tree layer colors alternate between black and red, with the top layer (containing the root) being black.

Essentially, this boils down to counting the number of internal nodes in a complete binary tree of height 2h - 1 (2h layers alternating between black and red). This is equal to the number of nodes in a complete binary tree of height 2h - 2 (since if you pull off all the leaves, you're left with a complete tree of height one less than what you started with). This works out to 2^{2h - 1} - 1, which differs from the number that you were given (which I'm now convinced is incorrect) but matches the number that you're getting.